common base bjt amplifier common collector bjt...
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ESE319 Introduction to Microelectronics
12009 Kenneth R. Laker, updated 05Oct12 KRL
Common Base BJT AmplifierCommon Collector BJT Amplifier
● Common Collector (Emitter Follower) Configuration● Common Base Configuration● Small Signal Analysis● Design Example● Amplifier Input and Output Impedances
ESE319 Introduction to Microelectronics
22009 Kenneth R. Laker, updated 05Oct12 KRL
Basic Single BJT Amplifier Features
CE Amplifier CC Amplifier CB AmplifierVoltage Gain (AV) moderate (-R
C/R
E) low (about 1) high
Current Gain (AI) moderate ( ) moderate ( ) low (about 1)
Input Resistance high high low
Output Resistance high low high
1
CE BJT amplifier => CS MOS amplifierCC BJT amplifier => CD MOS amplifierCB BJT amplifier => CG MOS amplifier
VCVS CCCS
ESE319 Introduction to Microelectronics
32009 Kenneth R. Laker, updated 05Oct12 KRL
Common Collector ( Emitter Follower) Amplifier
In the emitter follower, the output voltage is taken between emitter and ground. The voltage gain of this amplifier is nearly one – the output “follows” the input - hence the name: emitter “follower.”
vs
vO
RE
R1
R2
C inV CC vOv s
ro
Current Bias DesignVoltage Bias Design
ro
Q1 Q
2
Qamp
RS
ESE319 Introduction to Microelectronics
42009 Kenneth R. Laker, updated 05Oct12 KRL
Equivalent Circuits
<=>vout
vout
VCC
/2
Rb
RB=R1∥R2
RE
RB
C in
V Bvs
vORS
V CC
RE
vO
ESE319 Introduction to Microelectronics
52009 Kenneth R. Laker, updated 05Oct12 KRL
Multisim Bias Check
Identical results – as expected!
<=>Rb
+
-VRb
V Rb=I B RB=I E
1RB=0.495V
iB
Rbvout vout
Cin Cin
ESE319 Introduction to Microelectronics
62009 Kenneth R. Laker, updated 05Oct12 KRL
Follower Small Signal Analysis - Voltage GainCircuit analysis:
ib=vs
RSr1RE
vo=RE 1v s
RSr1RE
AV=vo
v s=
RE
RSr
1RE
≈1
vs=RSr ibRE ie=RSr1RE ib
vout
ib
ie
Solving for ib
vs
RS
RE
vo=ve
vo=RE ie=RE 1ib
for Current Bias Designreplace RE with ro||ro = ro/2 >> RE
ro∥ro
ro∥ro
ESE319 Introduction to Microelectronics
72009 Kenneth R. Laker, updated 05Oct12 KRL
Small Signal Analysis – Voltage Gain - cont.vo
v s=
RE
RSr
1RE
Since, typically:
RSr
1≪RE
AV=vo
v s≈
RE
RE=1
Note: AV is non-inverting
(or ro||ro = ro/2)
ESE319 Introduction to Microelectronics
82009 Kenneth R. Laker, updated 05Oct12 KRL
Quick Review CE Amplifier CC Amplifier CB Amplifier
Voltage Gain (AV)
Current Gain (AI)
Input Resistance
Output Resistance
ANSWERS: Low, Moderate or High
ESE319 Introduction to Microelectronics
92009 Kenneth R. Laker, updated 05Oct12 KRL
Quick Review cont. CE Amplifier CC Amplifier CB Amplifier
Voltage Gain (AV) moderate (-RC/R
E) low (about 1) high (R
C/R
S)
Current Gain (AI) moderate (β) moderate (β + 1) low (about 1)
Input Resistance high (RB||βR
E) high (R
B||βR
E) low (r
e)
Output Resistance high (RC||r
o) low (r
e) high (R
C||r
o)
VCVS CCCS
ESE319 Introduction to Microelectronics
102009 Kenneth R. Laker, updated 05Oct12 KRL
Of What value is a Unity Gain Amplifier?
To answer this question, we must examine the small-signal output impedance of the amplifier and its power gain.
ib
ie
vs vo
RE
RS
ESE319 Introduction to Microelectronics
112009 Kenneth R. Laker, updated 05Oct12 KRL
Emitter Follower Output Resistance
vx
ix
Rout
0ib
i x=−ie=−1 ib⇒ ib=−i x
1
vx=−ibRSr=RSr
1ix
Assume:I C=1mA⇒ r=
V T
I B=
V T
I C=2500
=100 RS=50
Rout≈re=2550100
=25.5
vs
RS
Rin=rbg=r1RERecall
ie
Rout=v x
i x=
RSr
1≈
r
1=r e=
1g m
=V T
I C
r≫RSwhere
ESE319 Introduction to Microelectronics
122009 Kenneth R. Laker, updated 05Oct12 KRL
Multisim Verification of Rout
Multisim short circuit check( = 100, vo = vs):
Rout=voc
isc=
AV vs rms
isc rms = 1
0.0396=25.25
Thevenin equivalent for the short-circuited emitter follower.
Rout
Av*vsigAV = 1 <=>
isc=i x
isc=i x
Rin
i x=−1ib
vsig=RS ibr ib
Rout=AV v sig
i x=
RSr
1Rin=RSr1RE≈1RE
+
-voc=AV vs
vs
RS
=100
vs
If β = 200, as for most good NPN transistors, Rout would be lower - close to 12 Ω.
Rout=voc
isc
Rin=rbg=RSr1RE≈1RE
v s=RS ibr ib
Rout=v s
isc=
RSr
1Routv oc=v s
Rout=voc
i sc=
vs rms
isc rms =
1V39.61 mA
=22.25
ESE319 Introduction to Microelectronics
132009 Kenneth R. Laker, updated 05Oct12 KRL
Emitter Follower Power GainConsider the case where a R
L=50 Ω load is connected
through an infinite capacitor to the emitter of the follower.Using its Thevenin equivalent:
vo=RL AV v s
RL∥RERout=
5075
v s=23
vs
io=AV v s
RoutRL∥RE=
v s
75
po=vo io=2
225vs
2
p s=vs is≈1
5000v s
2
50 load is in parallel with 5.1k RE and dominates:
Apwr=po
p s=
25000225
=44.4≫1
RE∥RL=5.1 k ∥50≈50
@ midband where Av = 1 is=ib=
vs
Rin≈
v s
1RL∥RE≈
v s
101⋅50≈
v s
5000
+-
-vth=G vsig
Rout≈25
RE∥RL=50+
C=∞is
vs voAv v s
ioRin
vbg
+
-
Rin
Rb
ibi
b
ie
RS
vbgRE
RBvs
C in
vo
rbg
Rin
R L
ESE319 Introduction to Microelectronics
142009 Kenneth R. Laker, updated 05Oct12 KRL
R1=R2
For an assumed = 100:
RB=R1∥R2=R1
2=1
10RE≈10 RE
R1=R2=20 REV B=V CC
2
RE=V E
I E=
V CC /2−0.7I E
⇒
Then, choose/specified IE, and the rest of the design follows:
Vb
iB
iC
iE
vout
As with CE bias design, stable op. pt. => RB≪1RE , i.e.
Emitter Follower Biasing – Typical Design
Split bias voltage drops aboutequally across the transistorV
CE (or V
CB) and VRe (or VB).
For simplicity,choose:
V B
RE
V CC
R1
R2
C in
RS
vs
vO=vE
ESE319 Introduction to Microelectronics
152009 Kenneth R. Laker, updated 05Oct12 KRL
Typical Design - Cont.Given: Rout=r e=25
V CC=12VAnd the rest of the design follows:
RE=V E
I E=12/2−0.7
10−3 =5.3 k
Use standard sizes
U
R1=R2=100 k
RE=5.1 k
5.1 kΩ
100 kΩ
100 kΩ
12V
C in
vO=vE
I E≈ I C=V T
re=1 mA
ESE319 Introduction to Microelectronics
162009 Kenneth R. Laker, updated 05Oct12 KRL
ib=vbg
r1RE
Use the base current expression:
To obtain the base to ground resistance of the transistor:This transistor input resistance is in parallel with R
B = 50 k ,
forming the total amplifier input resistance:
Rin=RSRB∥rbg≈RB∥rbg=515
5155050 k =45.6 k≈RB=50 k
vbg=r ibRE iE=r1RE ib
rbg=vbg
ib=r1RE≈1RE=101⋅5.1 k=515 k
RS=50
vbg
+
-
Rin
Rb
ibi
b
ie
RS
vbg RERB
vs
C in
vo
rbg
Rin
Blocking Capacitor - Cin - Selection
ESE319 Introduction to Microelectronics
172009 Kenneth R. Laker, updated 05Oct12 KRL
Cin – Selection cont.
RB=50 k
Assume fmin = 20 Hz
C in ≥10
2⋅20⋅50⋅103=1.59F
Choose Cin such that its reactance is ≤ 1/10 of RB at fmin:
C in ≥10
2 f min RB
12 f C in
=RB
10
Pick Cin = 3.3 µ F ), the nearest standard value in the Detkin Lab. We could be (unnecessarily) more precise and include r
bg and Rs
as part of the total resistance in the loop.
with
ESE319 Introduction to Microelectronics
182009 Kenneth R. Laker, updated 05Oct12 KRL
Final Design
3.3 uF
vs
vo
C in
RE
R1
R2
RS
V CC
ESE319 Introduction to Microelectronics
192009 Kenneth R. Laker, updated 05Oct12 KRL
Multisim Simulation Results
20 Hz Data
1 kHz Data
Av = 0.995
ESE319 Introduction to Microelectronics
202009 Kenneth R. Laker, updated 05Oct12 KRL
The Common Base Amplifier
Voltage Bias Design Current Bias Design
ESE319 Introduction to Microelectronics
212009 Kenneth R. Laker, updated 05Oct12 KRL
Common Base ConfigurationBoth voltage and current biasing follow the same rules asthose applied to the common emitter amplifier.
As before, insert a blocking capacitor in the input signal pathto avoid disturbing the dc bias.
The common base amplifier uses a bypass capacitor – or adirect connection from base to ground to hold the base atground for the signal only!
RECALL: The common emitter amplifier (except for intentional REfeedback) holds the emitter at signal ground, while the commoncollector circuit does the same for the collector.
ESE319 Introduction to Microelectronics
222009 Kenneth R. Laker, updated 05Oct12 KRL
We keep the same bias that we established for the gain of 10 common emitter amplifier.
All that we need to do is pick the capacitor values and calculate the circuit gain.
Voltage Bias Common Base DesignCE Amp
ESE319 Introduction to Microelectronics
232009 Kenneth R. Laker, updated 05Oct12 KRL
Mid-band Small Signal AnalysisvRe=r e∥RE is
Z in=v Re
i s=re∥RE≈re=
V T
I C
RC
Ai=ic
ie=≈1
Input Impedance
i s
Zin
Current Gain
Voltage Gain
vo=−RC ic=− RC ie=1
RC
RSre∥REv s
Av=vo
vs≈ 1
RC
RSre
Note: is = - i
eR
E >>r
e
ro
v s=−ie RS−re∥RE ie
C in=∞
Z out=vo
ic=RC∥ro
ESE319 Introduction to Microelectronics
242009 Kenneth R. Laker, updated 05Oct12 KRL
Common Base Small Signal Analysis - Cin
Determine Cin:4.7 k Ohm
470 Ohmideally for f ≥ f min
12 f min C in
≪RSRE∥r e⇒1
2 f min C in=
RSre
10⇒C in=
102 f min RSre
vRe=RE∥r e
RE∥reRS1
j2 f C in
v s
vRe=RE∥re
RE∥reRSvs
(let ) Cb=∞∞
re=r
1
re
v sNOTE:RB is shorted by Cb = ∞
vRe
RB
re
25 Ohm
RE
RS
Cb
Cb
ibR
C
ic
ie
vo
C in
ESE319 Introduction to Microelectronics
252009 Kenneth R. Laker, updated 05Oct12 KRL
Determine CIN cont.
2 f min C inRSre≫1⇒C in≥10
2 f min RSre=
10220⋅75
F
A suitable value for Cin for a 20 Hz f
min with r
e = 25 Ω and R
S = 50 Ω:
C in=10
125.6⋅75≈1062F ! Not Practical!
Must choose smaller value of Cin.Choose: 2 f min C in RSre=1
C in=1
125.6⋅75≈106.2F
ESE319 Introduction to Microelectronics
262009 Kenneth R. Laker, updated 05Oct12 KRL
Small-signal Analysis - Cb
Note the ac reference currentreversals (due to v
s polarity)!
Determine Cb: (let )C in=∞
ignore RB
vRe=RE∥r e
1j 2 f C b1
RE∥r e1
j 2 f C b1RS
vs
ideally vRe=RE∥re
RE∥reRSvs for f ≥ f min
12 f minC b1
≪re⇒1
2 f minC b1=
r e
10⇒Cb=
102 f min r e 1
ESE319 Introduction to Microelectronics
272009 Kenneth R. Laker, updated 05Oct12 KRL
Determine - CB cont.i'b i'c
i'e
Choose (conservatively):vout
Cb=10
220 10025 =31.8F
i.e.
ib'
v s
Cb=10
2 f min 1re F
for fmin = 20 Hz
ignore RB
RE∥re≈re
4.7 k Ohm
vo
is
RC
ESE319 Introduction to Microelectronics
282009 Kenneth R. Laker, updated 05Oct12 KRL
Multisim Simulation
1060 uF
v s
31.8 uF vO
470 Ohm
4.7 k Ohm
AV=vo
vs= 1
RC
RSre≈
RC
RSre= 4700
5025=62.7
ESE319 Introduction to Microelectronics
292009 Kenneth R. Laker, updated 05Oct12 KRL
Multisim Frequency Response
20 Hz response
1 kHz Response Av sim =63.3Av theory =62.7
vertical axis is a linear scale
1 %